Problem: $g(x)=\sin^2(2x^3+3x)$ Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(12x^3+6x)(2x^3+3x)^2\sin(x)\cos(x)$ (Choice B) B $(12x^2+6)\sin(2x^3+3x)\cos(2x^3+3x)$ (Choice C) C $2\sin(2x^3+3x)\cos(2x^3+3x)$ (Choice D) D $(12x^2+3)(2x^3+3x)\cos((2x^3+3x)^2)$
Answer: $\sin^2(2x^3+3x)$ is a composition of three functions! Let... $u(x)=x^2$ $v(x)=\sin(x)$ $w(x)=2x^3+3x$... then $g(x)=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $g'(x)$, we will need to use the chain rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=2x$ $v'(x)=\cos(x)$ $w'(x)=6x^2+3$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={2\sin(2x^3+3x)}\cdot{ \cos(2x^3+3x)}\cdot{(6x^2+3)} \\\\ &=(12x^2+6)\sin(2x^3+3x)\cos(2x^3+3x) \end{aligned}$ In conclusion: $g'(x)=(12x^2+6)\sin(2x^3+3x)\cos(2x^3+3x)$